How do you solve #ln((e^(4x+3))/e)=1#?

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Answer 1 · 2015-12-15T09:52:09

#x = -1/4#

Use the following logarithmic law first:

#ln (a/b) = ln(a) - ln(b)#

In your case, this leads to:

#ln(e^(4x+3)/e) = 1#

#<=> ln(e^(4x+3)) - ln(e) = 1#

As next, you need to use the property that #ln x# and #e^x# are inverse functions which means that #ln(e^x) = x# and #e^(ln x) = x# always hold.

Thus, #ln# and #e# eliminate each other in your equation, and you will get:

#<=> (4x + 3) - 1 = 1#

The solution of this equation is

#x = -1/4#

As #e^x# is always positive for any value of #x in RR#, and thus the logarithmic expression is defined for any #x in RR#, this is your solution.